Solving exponential equations like $2^{2x} - 3 \cdot 2^x - 10 = 0$ I have two equations that I'm not able to solve. I know the answers, but I can't get to them.

$$(a) \qquad 2^{2x} - 3 \cdot 2^x - 10 = 0$$
(Answer: $x = \frac{\log 5}{\log 2}$.)

On a) I start by multiplying log into all of the numbers. But then I realized that I might be misunderstanding. How will the middle part look, from what I have done I get ...$(-\log3 + x\log2)$... Am I supposed to split 3*2 apart(following the rule: $\log(a b)=\log a+\log b)$? And if so, will there be a - or a + in front of number $2$? The answer that I've got that is closest is ${(\log(10)+\log(3)) / \log(2)}$.

$$(b) \qquad 3^{2x} - 12 \cdot 3^x + 27 = 0$$
(Answer: $x = 1, 2$.)

On b) I don't have anything close to the answer. I understand that the quadratic equation formula is needed, but I can't get to those numbers that I'm supposed put in it.
Does anyone know how to solve this, and what I'm doing wrong?
 A: $(a)\;\; $ Given $2^{2x}-3\cdot 2^{2x}-10 =0\;\;,$ Now Let $2^x=y>0\forall y\in \mathbb{R}\;,$ Then equation convert into
$$\implies y^2-3y-10=0\Rightarrow y=\frac{3\pm \sqrt{9+40}}{2} = \frac{3\pm 7}{2}$$
So we get $$y=5\Rightarrow 2^x=5\Rightarrow \log_{2}(2)^x=\log_{2}(5)\Rightarrow x=\log_{2}(5)\;\;$$ and $$y=-4\;\bf{(Not\; Possible)}$$
$(b)$ Given $3^{2x}-12\cdot 3^x+27=0\;,$ Put $3^x=y>0\forall y\in \mathbb{R}$
Same process as above.
A: To see that $a^{2x}=(a^x)^2$ might be the most important thing here.
Then, you can set $2^x=s$ for a) (be careful about $s\gt 0$), and $3^x=t$ for b).
A: We can regard the first equation as a quadratic equation in $u := 2^x$ and then factor:
$$u^2 - 3 u - 10 = (u + 2) (u - 5) = 0.$$
(Conveniently, factoring lets us bypass the quadratic equation here.)
We can see immediately that this equation has solutions $u = - 2$ and $u = 5$. To which values of $x$ do these solutions correspond, if any?
Part (b) can be handled similarly.
A: HINT: rewrite the first equation in the form
$(2^x)^2-3\cdot 2^x-10=0$ and set $2^x=t$
and the second one in the form
$(3^x)^2-12\cdot 3^x+27=0$
A: Take $2^x$ in first equation as $a$ and $3^x$ in 2nd equation as $y$. You will get $$a^2-3a-10=0$$ or, $$(a-5)(a+2)=0$$ and $$b^2-12b+27=0$$ or, $$(b-3)(b-9)=0$$ and these are easy quadratic equations to solve. 
A: Notice, we have
1. $$2^{2x}-3(2^x)-10=0\implies (2^x+2)(2^x-5)=0$$ 
$$2^x=-2, \ 5$$
but, $2^x$ is non-negative hence, we have $$2^x=5$$ $$\ln(2^x)=\ln(5)$$   $$x\ln(2)=\ln(5) \implies \color{red}{x=\frac{\ln 5}{\ln 2}}$$


*$$3^{2x}-12(3^x)+27=0\implies (3^x-9)(3^x-3)$$ 
$$\implies 3^x=3, 9$$
$$3^x=3\implies \ln(3^x)=\ln(3)$$ $$x\ln(3)=\ln(3)\implies \color{red}{x=1}$$
$$3^x=9\implies \ln(3^x)=\ln(9)$$$$x\ln(3)=2\ln(3)\implies \color{red}{x=2}$$

